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Excess of KI ...

Question

Excess of KI reacts with $C u S O_{4}$ solution and then $N a_{2} S_{2} O_{3}$ solution is added to it. Which of the following statement is incorrect from this reaction

A

$N a_{2} S_{2} O_{3}$ oxidized

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B

$C u_{2} I_{2}$ is formed

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C

$C u I_{2}$ is formed

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D

Evolved $I_{2}$ is reduced

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Solution

The correct option is B
$C u_{2} I_{2}$ is formed

$- 1 4 K I + + 2 2 C u S O_{4} \to 0 C u_{2} I_{2} + 2 K_{2} S O_{4}_{2} + 20 2 N a_{2} S_{2} O_{3} \to + 2.5 N a_{2} S_{4} O_{6} + - 1 2 N a I$

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Similar questions

K I

C u S O_{4}

solution and then

N a_{2} S_{2} O_{3}

solution is added to it. Which of the following statements is incorrect for this reaction?

C u S O_{4}

reacts with excess amount of

K I

followed by solution of

N a_{2} S_{2} O_{3}

. In this process which of the following statement is correct?

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

0.02 M

solution with excess

K I

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.

The color of the solution developed during the addition of

K I

C u S O_{4}

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.

Percentage of purity of sample is:

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